Proof that higher cech cohomology groups vanish for fine sheaves.

So I've been trying to understand a proof of the fact that $H^p(M, \mathcal F)=0$ whenever $p\geq 1$ from page 42 of the book "Principles of Algebraic Geometry" by Griffiths and Harris.

The proof is carried out for the sheaf ${\mathcal a}^{r,s}$ of $C^{\infty}$ forms of type $(r,s)$ on $U$ and there is a remark after the proof of how one can prove the result for an arbitrary fine sheaf.

According to them, a fine sheaf is a sheaf for which we have maps $\eta_\alpha:\mathcal F (U_{\alpha})\rightarrow \mathcal F (U)$ for any $U=\cup U_{\alpha}$, such that the support of $(\eta_{\alpha}\sigma)$ is contained in $U_\alpha$ and $\sum {\eta_{\alpha}} (\sigma| _{U_\alpha})=\sigma$ for $\sigma\in \mathcal F (U)$.

So is $(\eta_{\alpha}\sigma)$ a map from the manifold $M$ to $\mathbb R$? I don't understand how $(\eta_{\alpha}\sigma)$ has $M$ as its domain.

Secondly how can they write $(\eta_{\alpha}\sigma)$ when $\sigma\in \mathcal F (U)$? Isn't the domain of $\eta_\alpha$ the group ${\mathcal F }(U_\alpha)$?

Thirdly how does this formulation apply to the case when $\mathcal F$ is the sheaf ${\mathcal a}^{(r,s)}$? We would need maps $\eta_{\alpha}: {\mathcal a}^{(r,s)}(U_{\alpha})\rightarrow {\mathcal a}^{(r,s)}(U)$. But indstead the proof just uses the partion of unity $C^{\infty}$ maps $\rho_{\alpha}$ subordinate to some open cover $\{U_\alpha\}$. So it seems that they haven't used the fact that ${\mathcal a}^{r,s}$ is a fine sheaf.

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miss a "$" in the third paragraph? It doesn't display correctly here...And I think in order to imitate the partition of unity, G&H abuses the language here. Probably you need a reference for a more detailed version about these. I learned these stuff from Voisin's hodge theory and complex algebraic geometry. – Honglu May 27 '11 at 6:27 I've tried to fix up the latex - please correct if I've destroyed the meaning of what was intended. – David Roberts May 27 '11 at 7:20 Yes, thank you David. That's what I intended. – Nakhoul May 27 '11 at 7:27 Yes the problem with learning from Grifiths and Harris is that they constantly abuse notation, making a lot of what they say somewhat elusive. – Nakhoul May 27 '11 at 7:33 1 Answer About your third question. Using partition of unity$(\rho_\alpha)_\alpha$subordinate to$(U_\alpha)_\alpha$for$\mathcal F$you can define maps$\eta_\alpha:\mathcal F(U_\alpha)\to\mathcal F(U)$by taking$s\in\mathcal F(U_\alpha)$to$\rho_\alpha s$. About the first two points it seems to me that$(\eta_\alpha\sigma):=\eta_\alpha(\sigma_{|U_\alpha})$. EDIT: to summarize my answer a fine sheaf is, "by definition", a sheaf having partition of unity. - Thanks. Though I still don't understand what it means to say that$support(\eta_\alpha \sigma)\subset U_\alpha$. I can understand what it means in the case where$\mathcal F$is the sheaf of$C^\infty$functions, but when it comes to an arbitrary element$\sigma$of some section$\mathcal F (U)$, I cannot fathom what it means for$\sigma$to only be non zero on$U_\alpha$, where$U_\alpha$is some element of a open cover of$M$. – Nakhoul May 27 '11 at 13:28 For a section$\tau\in \mathcal F(U)$(e.g.$\tau=(\eta_\alpha\sigma)$), the complement$C$of its support is defined as follows: a point$x\in U$lies in$C$iff there exists an open neighborhood$V$of$x$in$U$such that$tau_{|V}=0\$. – DamienC May 27 '11 at 14:56
Thanks so much for clearing that up. – Nakhoul May 27 '11 at 15:34